anonymous
  • anonymous
I have the answer but i dont know how.. The lower edge of a mural, 12 ft high, is 6ft above an observer’s eye. Under the assumption that the most favorable view is obtained when the angle subtended by the mural at the eye is maximum, at what distance from the wall should the observer stand?
Mathematics
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anonymous
  • anonymous
I have the answer but i dont know how.. The lower edge of a mural, 12 ft high, is 6ft above an observer’s eye. Under the assumption that the most favorable view is obtained when the angle subtended by the mural at the eye is maximum, at what distance from the wall should the observer stand?
Mathematics
chestercat
  • chestercat
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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ganeshie8
  • ganeshie8
|dw:1433509816192:dw|
anonymous
  • anonymous
we have the same drawing sir
ganeshie8
  • ganeshie8
From the lower small triangle \[\tan(s) = \frac{6}{x}\tag{1}\] From the big triangle \[\tan(t+s)=\frac{18}{x}\tag{2}\]

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ganeshie8
  • ganeshie8
Our goal is to eliminate \(s\) somehow and get an equation with \(t\) and \(x\) as variables
ganeshie8
  • ganeshie8
In view of that, apply the angle sum formula : \[\large \tan(t)=\tan((t+s)-s)=\frac{\tan(t+s)-\tan(s)}{1+\tan(t+s)\tan(s)}\] plugin the values
anonymous
  • anonymous
ok sir i will solve now
anonymous
  • anonymous
i got this sir: |dw:1433510419375:dw|
ganeshie8
  • ganeshie8
doesn't look correct
ganeshie8
  • ganeshie8
\[\large{\begin{align} \tan(t)=\tan((t+s)-s) &=\frac{\tan(t+s)-\tan(s)}{1+\tan(t+s)\tan(s)}\\~\\ &=\dfrac{\dfrac{18}{x}-\dfrac{6}{x}}{1+\dfrac{18}{x}\dfrac{6}{x}}\\~\\ &=\dfrac{12x}{x^2+108} \end{align}}\]
ganeshie8
  • ganeshie8
so the equation is \[\large \tan(t)=\dfrac{12x}{x^2+108}\] and we want to maximize \(t\)
anonymous
  • anonymous
i got that equation sir
ganeshie8
  • ganeshie8
may be isolate \(t\) and find \(\dfrac{dt}{dx}\)
anonymous
  • anonymous
is this the bases sir? |dw:1433510949377:dw|
ganeshie8
  • ganeshie8
\[\large \tan(t)=\dfrac{12x}{x^2+108}\] \[\large t = \tan^{-1}\left(\dfrac{12x}{x^2+108}\right)\] \[\large \dfrac{dt}{dx} = ?\]
ganeshie8
  • ganeshie8
remember the derivative of \(\tan^{-1}(x)\) ?
anonymous
  • anonymous
|dw:1433511150272:dw|
ganeshie8
  • ganeshie8
Yes use that
ganeshie8
  • ganeshie8
\[\large t = \tan^{-1}\left(\dfrac{12x}{x^2+108}\right)\] \[\large{\begin{align} \dfrac{dt}{dx} &= \dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\dfrac{d}{dx}\left(\dfrac{12x}{x^2+108}\right)\\~\\~\\ &=? \end{align}}\]
anonymous
  • anonymous
ok wait sir i will solve
anonymous
  • anonymous
i got this sir:
anonymous
  • anonymous
\[\frac{ 1 }{ 1+\frac{ 12X ^{2} }{ x ^{4}+216x ^{2}+11664 } }\times \frac{ -12x ^{2}+1296 }{ x ^{4}+216x ^{2}+11664 }\]
anonymous
  • anonymous
ohh there an error im sorry.. it should be 144x^2 instead of having 12x^2
anonymous
  • anonymous
and i was stuck into this.. i dont know how to continue...
anonymous
  • anonymous
i already have an headache..
ganeshie8
  • ganeshie8
do not expand
ganeshie8
  • ganeshie8
because we will be setting all that equal to 0 and denominator gona vanish
ganeshie8
  • ganeshie8
\[\large{\begin{align} \dfrac{dt}{dx} &= \dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\dfrac{d}{dx}\left(\dfrac{12x}{x^2+108}\right)\\~\\~\\ &=\dfrac{1}{1+\left(\dfrac{12x}{x^2+108}\right)^2}*\left(\dfrac{(x^2+108)12-12x(2x)}{(x^2+108)^2}\right)\\~\\~\\ \end{align}}\] yes ?
ganeshie8
  • ganeshie8
setting that equal to 0 gives \[\large (x^2+108)12-12x(2x) = 0\] solve \(x\)
anonymous
  • anonymous
x=10.4
anonymous
  • anonymous
but sir im a little bit confused by this... how come the remaining equation becomes \[(x ^{2}+108)12-12x(2x)\]
anonymous
  • anonymous
how did the others vanish? and this is the equation that still remains?
anonymous
  • anonymous
where did this \[\frac{ 1 }{ 1 +\left( \frac{ 12x }{ x ^{2}+108 } \right)^{2}}\] go?

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